1. P460 - Relativity21 Lorentz Transformations (px,py,px,E) are components of a 4-vector which has same Lorentz transformation px’ =  (px + uE/c 2 ) u = velocity of transform py’ = py between frames is in pz’ = pz x-direction. If do px’ -> px E’ =  (E + upx) + -> - Use common sense also let c = 1 Frame 1 Frame 2 (cm) Before and after scatter

2. P460 - Relativity22 center-of-momentum frame  p = 0. Some quantities are invariant when going from one frame to another: py and pz are “transverse” momentum M total = Invariant mass of system dervived from Etotal and Ptotal as if just one particle How to get to CM system? Think as if 1 particle Etotal = E1 + E2 Pxtotal = px1 + px2 (etc) Mtotal 2 = Etotal 2 - Ptotal 2  cm = Etotal/Ptotal and  cm = Ptotal/Etotal Lab CM 1 2 at restp1 = p 2 1212

3. P460 - Relativity23 Particle production convert kinetic energy into mass - new particles assume 2 particles 1 and 2 both mass = m Lab or fixed target Etotal = E1 + E2 = E1 + m2 Ptotal = p1 --> Mtotal 2 = Etotal 2 - Ptotal 2  Mtotal = (E1 2 +2E1*m + m*m - p1 2 ).5 Mtotal= (2m*m + 2E1*m) 0.5 ~ (2E1*m) 0.5 CM: E1 = E2 Etotal = E1+E2 and Ptotal = 0 Mtotal = 2E1 Lab CM 1 2 at restp1 = p 2 1212

4. P460 - Relativity24 p + p --> p + p + p + pbar what is the minimum energy to make a proton- antiproton pair? In all frames Mtotal (invariant mass) at threshold is equal to 4*mp (think of cm frame, all at rest) Lab Mtotal = (E1 2 +2E1*m + m*m - p1 2 ).5 Mtotal= (2m*m + 2E1*m) 0.5 = 4m E1 = (16*mp*mp - 2mp*mp)/2mp = 7mp CM: Mtotal = 2E1 = 4mp or E1 and E2 each =2mp Lab CM 1 2 at restp1 = p 2 1212

5. P460 - Relativity25 Transform examples Trivial: at rest E = m p=0. “boost” velocity = v E’ =  (E +  p) =  m p’ =  p +  E) =  m moving with velocity v = p/E and then boost velocity = u (letting c=1) E’ =  (E + pu) p’ =  p + Eu) calculate v’ = p’/E’ = (p +Eu)/(E+pu) = (p/E + u)/(1+up/E) = (v+u)/(1+vu) “prove” velocity addition formula

6. P460 - Relativity26 A p=1 GeV proton hits an electron at rest. What is the maximum pt and E of the electron after the reaction? Elastic collision. In cm frame, the energy and momentum before/after collision are the same. Direction changes. 90 deg = max pt 180 deg = max energy  cm = Ptot/Etot = Pp/(Ep + me) ·Pcm =  cm  cm*me (transform electron to cm) ·Ecme =  cm*me (“easy” as at rest in lab) · pt max = Pcm as elastic scatter same pt in lab Emax =  cm(Ecm +  cmPcm) 180 deg scatter =  cm(  cm*me +  *me) =  *me(1 +  )

7. P460 - Relativity27 p=1 GeV proton (or electron) hits a stationary electron (or proton) mp =.94 GeV me =.5 MeV incoming target  cm  cm Ptmax Emax p e.7 1.5.4 MeV 1.7 MeV p p.4 1.2.4 GeV 1.4 GeV e p.5 1.2.5 GeV 1.7 GeV e e.9995 30 15 MeV 1 GeV  max is maximum energy transferred to stationary particle. Ptmax is maximum momentum of (either) outgoing particle transverse to beam. Ptmax gives you the maximum scattering angle a proton can’t transfer much energy to the electron as need to conserve E and P. An electron scattering off another electron can’t have much Pt as need to conserve E and P.